Objectives
Degree 2 polynomials are quadratic polynomials.
ax
2
+ bx + c where a , b , c ∈
R and a = 0
The graph of a quadratic polynomial is called a parabola. The vertex is a feature of a parabola. Identifying this feature helps to graph the parabola.
We are also interested in roots of these polynomials. To summarize:
I Completing the square identifies the vertex.
I The discriminant identifies the number of roots.
I The quadratic formula idenitifies the value of roots, if any.
Each of these items must be memorized.
(University of Utah) Math 1050 1 / 6
Notes
Vertex
Recall graph transformations and the point (0 , 0) on the graph of f ( x ) = x
2
. This is its vertex.
We shift horizontally by β ∈
R
, then f ( x + β ) = ( x + β )
2
(0 , 0) to the point ( − β, 0).
moves the point
Additionally we can vertically scale and also reflect the graph by multiplying by α ∈
R
, and we can shift vertically by adding δ ∈
Together with a horizontal shift gives
R
.
α f ( x + β ) + δ = α ( x + β )
2
+ δ.
This moves the vertex (0 , 0) on the graph of f to the point ( − β, δ ).
(University of Utah) Math 1050 2 / 6
Notes
Parabola and its vertex
(1 , 3) f ( x ) = x
2 h ( x ) = ( x + 3)
2
( − 3 , 0) (0 , 0) g ( x ) = − 2( x − 1)
2
+ 3
(University of Utah) Math 1050 3 / 6
Notes

Completing the Square
We have a quadratic polynomial ax
2
+ bx + c where a , b , c ∈
R and a = 0
What is the vertex?
Completing the Square (Memorize) ax
2 b
+ bx + c = a x +
2 a
The vertex is the point − b
2 a
, c − b
2
4 a
.
2 b
2
+ c −
4 a
Example If p ( x ) = 5 x
2
+ 40 x + 81, then completing the square shows us b
2 a
=
40
10
= 4 and c − b 2
4 a
= 81 −
1600
20
= 81 − p ( x ) = 5( x + 4)
2
+ 1 has a vertex at ( − 4 , 1).
80 = 1 and thus
(University of Utah) Math 1050 4 / 6
Notes
How many roots?
ax
2
+ bx + c where a , b , c ∈
R and a = 0
A degree 2 polynomials has at most 2 roots. (0, 1, and 2 are possibilities).
The discriminant of a quadratic polynomials is the number b
2 − 4 ac .
b
2 − 4 ac Number of Roots
> 0
= 0
< 0
2 roots
1 root
0 roots
Example g ( x ) = − 2 x
2 b
2 − 4 ac = (4)
2 − 4( −
+ 4 x + 1 has discriminant
2)(1) = 16 + 8 = 24 .
This number is positive (that is all that matters). Then g ( x ) has 2 roots.
(University of Utah) Math 1050 5 / 6
Notes
What are the value(s) of the roots, if any?
ax
2
+ bx + c where a , b , c ∈
R and a = 0
The quadratic formula will tell us the value or values of the roots whenever they exist.
Quadratic formula (Memorize)
If p ( x ) is a quadratic polynomial with roots (that is b
2
− 4 ac ≥ 0), then the roots have value
− b +
√ b 2 − 4 ac and
− b −
√ b 2 − 4 ac
2 a 2 a
Example g ( x ) = − 2 x
2
− 4 + p
4 2 − 4( − 2)(1)
− 4 − p
4
2
− 4
−
−
4
4( − 2)(1)
+ 4 x + 1 has two roots. They are
=
− 4 + 24
=
− 4 + 2
√
6
=
2 −
=
− 4 √
− 4 − 24
=
− 4
− 4 − 2
√
6
=
√
6
2 +
2 √
6
− 4 − 4 2
(University of Utah) Math 1050 and
6 / 6
Notes